Question

3. Use the following equation (#6): 2CH2 + 170, → 12 CO2 + 10H0 If you have 2.67 g of the organic fuel, how many grams of carbon dioxide will you produce? 4. Use the following equation (#7): 2C,H,02 + 3 02 → 400 + 6 H,0 If you have 5.88 g of oxygen gas, how many grams of water will you produce?

Answer 1

3)

Molar mass of C6H10,

MM = 6*MM(C) + 10*MM(H)

= 6*12.01 + 10*1.008

= 82.14 g/mol

mass of C6H10 = 2.67 g

mol of C6H10 = (mass)/(molar mass)

= 2.67/82.14

= 3.251*10^-2 mol

According to balanced equation

mol of CO2 formed = (12/2)* moles of C6H10

= (12/2)*3.251*10^-2

= 0.195 mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 0.195*44.01

= 8.583 g

Answer: 8.58 g

4)

Molar mass of O2 = 32 g/mol

mass of O2 = 5.88 g

mol of O2 = (mass)/(molar mass)

= 5.88/32

= 0.1837 mol

According to balanced equation

mol of CO2 formed = (6/3)* moles of O2

= (6/3)*0.1837

= 0.3675 mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 0.3675*44.01

= 16.17 g

Answer: 16.2 g